M408M Learning Module Pages

Main page

Tangent planes

Tangent Planes Tangent Plane: to determine the equation of the tangent plane to the graph of $z = f(x,\, y)$, let $P = (a,\,b,\, f(a,\,b))$ be a point on the surface above $(a,\,b)$ in the $xy$-plane as shown to the right below . Slicing the surface with vertical planes $y=b$ and $x=a$ creates two curves on this graph, both passing through $P$.

These two space curves are the graphs of vector functions $${\bf r}_1(x)\ = \ \big\langle\, x, \,b, \,f(x, b)\big \rangle\,, $$ $${\bf r}_2(y)\ = \ \big\langle \,a,\, y, \,f(a, y)\big \rangle,$$ shown in orange on the surface. The vector derivatives $${\bf T}_1 \ = \ {\bf r}_1'(a)\ = \ {\bf i}+ f_x(a, b)\,{\bf k}\,,$$ $${\bf T}_2\ = \ {\bf r}_2 '(b)\ = \ {\bf j}+ f_y(a, b)\,{\bf k}\,,$$ are Tangent Vectors at $P$ to the graph of $f$, while the plane containing $P$ as well as ${\bf T}_1$and ${\bf T}_2$ is the Tangent Plane at $P$.
To calculate the equation of the tangent plane at $P(a,\,b,\,c)$ we need its normal: $${\bf n} \ = \ {\bf T}_1 \, \times \, {\bf T}_2 \ = \ \left[\begin{array} {ccc} \,{\bf i} \,&\, {\bf j}\,& {\bf k } \\ 1 & 0 & f_x(a,\,b) \\ 0 & 1 & f_y(a,\,b) \end{array}\right]$$ $$\qquad = \ \big \langle\, - f_x(a,\,b),\, - f_y(a,\,b),\,1\, \big \rangle\,.$$

On the other hand, if $Q(x, \,y,\, z)$ is an arbitrary point in the tangent plane at $P(a,\,b,\,c)$, then $$\overrightarrow{PQ}\ = \ \big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle$$ lies in the plane and so is perpendicular to ${\bf n}$. Thus in point-normal form the equation of the tangent plane is $$\big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle \cdot {\bf n} \ = \ \big \langle\, x - a,\, y - b,\, z - f(a, b)\, \big \rangle \cdot \big \langle\, - f_x(a,\,b),\, - f_y(a,\,b),\,1\, \big \rangle$$ $$ = \ -(x-a)f_x(a, b) - (y-b)f_y(a, b) + z - f(a, b) \ = \ 0\,. \qquad \qquad$$ After rearranging we get:

The equation of the Tangent Plane at $(a,\, b,\, f(a, \,b))$ is $$z\ = \ f(a, b) + (x-a)f_x(a, b) + (y-b)f_y(a, b)\,.$$

Example: Determine the tangent plane to the graph of $$f(x,\, y) \ = \ x^3 +y^2 +2x$$ at $(-1,\,2, f(-1,2))$.
Solution: since $$\frac{\partial f}{\partial x} \,=\, 3x^2 +2,\quad \frac{\partial f}{\partial y} \,=\, 2y\,,$$

we see that $$f_x\Big|_{(-1,2)} \ = \ 5\,, \quad f_y\Big|_{(-1,2)} \ = \ 4\,.$$ Thus the tangent plane at $(-1,\,2, f(-1,2))$ is $$z \ = \ f(-1,\,2) +5(x+1) +4(y-2)\,,$$ which after simplification becomes $$z \ = \ 5x + 4y -2\,.$$

Chapter 10: Parametric Equations and Polar Coordinates

Chapter 12: Vectors and the Geometry of Space

Chapter 13: Vector Functions

Chapter 14: Partial Derivatives

Learning module LM 14.1: Functions of 2 or 3 variables:

Learning module LM 14.3: Partial derivatives:

Learning module LM 14.4: Tangent planes and linear approximations:

Learning module LM 14.5: Differentiability and the chain rule:

Learning module LM 14.6: Gradients and directional derivatives:

Learning module LM 14.7: Local maxima and minima:

Learning module LM 14.8: Absolute maxima and Lagrange multipliers:

Chapter 15: Multiple Integrals

Tangent planes